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Chapter 6 Exercises 1
1. Consider the reaction 2 AlCl3(s)  2 Al(s) + 3 Cl2(g) ΔHrxn = 139.8 kJ
Calculate q, w, and ΔE if 5.00 moles of AlCl3 are converted to products against a pressure of
1.00 atm at 75°C.
q = 5.00 mol AlCl3 x 139.8 kJ = 350. kJ
2 mol AlCl3
5.00 mol AlCl3 x 3 mol Cl2 = 7.50 mol Cl2
Δn(gases) = 7.50 mol – 0 = 7.50 mol
2 mol AlCl3
ΔV = ΔnRT/P = (7.50 mol)(0.0821 L atm/mol K)(348 K) / (1.00 atm) = 214 L
w = –PΔV = –(101.3 kPa)(214 L) = –2.17 x 104 J = –21.7 kJ
ΔE = q + w = 350. kJ – 21.7 kJ = 328 kJ
2.
CaO + 3 C → CaC2 + CO
H = +464 kJ/mol
Is this reaction exothermic or endothermic? When 375 kJ of heat are used, how many grams of
CaC2 are produced?
375 kJ x 1 mol CaC2 x 64.10 g CaC2 = 51.8 g CaC2
464 kJ
1 mol CaC2
3. A welded layer of copper on the bottom of a skillet weighs 125 g. How much heat is needed
to raise the temperature of the copper layer from 25.0ºC to 100.0ºC? The specific heat capacity
of Cu is 0.387 J/gK.
H = (125 g)(0.387 J/gK)(100.0 – 25.0°C)
= +3630 J or +3.63 kJ
4. 29.7 J of heat are transferred when a 5.5 g iron nail is cooled from 37ºC to 25ºC. Calculate
the specific heat capacity of iron.
C = H =
–29.7 J
= 0.45 J/g°C
m T
(5.5 g)(25 - 37°C)
(over)
5. A 25.64 g sample of a solid was heated in a test tube to 100.0ºC in boiling water and carefully
added to a coffee cup calorimeter containing 50.0 g water. The water temperature increased
from 25.10ºC to 28.49ºC. What is the specific heat capacity of the solid?
Hwater = (50.0 g)(4.18 J/g°C)(28.49 – 25.10°C) = +709 J
Hsolid = –709 J
C = H =
–709 J
= 0.386 J/g°C
m T
(25.64 g)(28.49 – 100.0°C)
6. In a coffee-cup calorimeter, 25.0 mL of 1.50 M HBr solution are mixed with 50.0 mL of
1.50 M KOH solution. Both solutions are initially at 20.0°C. The temperature of the combined
solution rises to 28.3°C. Assuming that the heat capacity of the solution is 4.18 J/g°C and the
density is 1.00 g/mL, calculate the enthalpy change (H) for the reaction in kJ/mol.
HBr + KOH  KBr + H2O
Hwater = (75.0 g)(4.18 J/g°C)(28.3 – 20.0°C) = +2600 J
Hrxn = –2600 J = –2.6 kJ
mol HBr = 0.0250 L x 1.50 mol/L = 0.0375 mol
divide kJ by moles:
–2.6 kJ = –69 kJ/mol
0.0375 mol
7. When a 6.50 g sample of solid sodium hydroxide dissolves in 100.0 mL of water in a coffee
cup calorimeter, the temperature rises from 21.6ºC to 37.8ºC. Calculate the change in enthalpy
in kJ, then calculate the enthalpy change for the reaction in kJ/mol. (Hint: the mass of the
solution is the sum of grams of water + grams of NaOH.)
NaOH(s)  Na+(aq) + OH–(aq)
Hwater = (106.5 g)(4.18 J/g°C)(37.8 – 21.6°C) = +7210 J = +7.21 kJ
Hrxn = –7.21 kJ
6.50 g NaOH x 1 mol NaOH x 0.163 mol
40.00 g NaOH
–7.21 kJ / 0.163 mol = –44.4 kJ/mol